How am I supposed to approach this problem? the points are A(0, 0), B(6, 0) and C (4, 4)
Plot the three points and draw the triangle.assuming you want the circumscribed circle erect the perpendicular bisectors of two sides.use 0,0 and 6,0 for one of them.from slope diagrams find the slope of each line (side). then find the center of each line. write two equations in x and y for linesperpendiculartoeach(point slope formula) at the midpoints. Solve the two equations for x and y the center of the circle.finallydetermine the
radiusby distance formular between center and a vertex.if this makes sense and you can proceed repost to write the circle equationor if you need more help
bjh
The general equation of a circle is
$\displaystyle (x - h)^2 + (y - k)^2 = r^2$
where $\displaystyle r$ is the radius and $\displaystyle (h, k)$ is the centre.
You know that $\displaystyle (0, 0), (6, 0), (4, 4)$ are three points on the circle. So you can get three equations in three unknowns:
$\displaystyle (0 - h)^2 + (0 - k)^2 = r^2$
$\displaystyle (6 - h)^2 + (0 - k)^2 = r^2$
$\displaystyle (4 - h)^2 + (4 - k)^2 = r^2$.
Equations 1 and 2 are equal, so
$\displaystyle (-h)^2 + (-k)^2 = (6 - h)^2 + (-k)^2$
$\displaystyle (-h)^2 = (6 - h)^2$
$\displaystyle h^2 = 36 - 12h + h^2$
$\displaystyle 0 = 36 - 12h$
$\displaystyle 12h = 36$
$\displaystyle h = 3$.
Substituting into the second and third equations gives:
$\displaystyle (6 - 3)^2 + (0 - k)^2 = r^2$ and $\displaystyle (4 - 3)^2 + (4- k)^2 = r^2$.
Therefore $\displaystyle 9 + k^2 = 1 + (4 - k)^2$
$\displaystyle 8 + k^2 = 16 - 8k + k^2$
$\displaystyle 8k = 8$
$\displaystyle k = 1$.
Substituting into the third equation:
$\displaystyle (4 - 3)^2 + (4 - 1)^2 = r^2$
$\displaystyle 1 + 9 = r^2$
$\displaystyle 10 = r^2$
$\displaystyle r = \sqrt{10}$.
So finally, the equation of the circle is
$\displaystyle (x - 3)^2 + (y - 1)^2 = (\sqrt{10})^2$.
Hi ziyadbasheer,
As (0,0) and (6,0) are two points on the circle,
the circle's centre lies on the line that is the perpendicular bisector
of the line segment starting at (0,0) and ending at (6,0).
This line segment is a chord of the circle and lies on the x-axis.
Hence the perpendicular bisector is vertical (parallel to the y-axis)
and passes through (3,0) which is the midpoint of the line segment from (0,0) to (6,0).
The circle centre also lies on the perpendicular bisector of the line segment
(which is another chord of the circle) going from (0,0) to (4,4).
The midpoint of this line segment is (2,2) and since the line has a 45 degree angle with the x-axis, going upwards from (0,0) to (4,4)
then the perpendicular bisector goes downwards at 45 degrees from (2,2)
and hence it crosses the x-axis at (4,0).
For this line, as x increases by 1, y decreases by 1.
Hence when it reaches x=3, (an increase of 1 in x) it will have decreased by 1
in the y direction.
Therefore the two perpendicular bisectors cross at (3,1).
This is the circle centre.
You can now draw the circle.
It's radius can be calculated with Pythagoras' theorem if you like
$\displaystyle 1^2+3^2=r^2=1+9=10$
$\displaystyle r=\sqrt{10}$